Difference between revisions of "2023 AMC 8 Problems/Problem 17"
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The answer is <math>\boxed{\textbf{(A)}\ 1}.</math> Use intuition to bring it down to <math>2</math> guesses <math>1</math> or <math>2</math> and guess from there or you could actually fold the paper. | The answer is <math>\boxed{\textbf{(A)}\ 1}.</math> Use intuition to bring it down to <math>2</math> guesses <math>1</math> or <math>2</math> and guess from there or you could actually fold the paper. | ||
-apex304 | -apex304 | ||
+ | |||
+ | ==Solution 2== | ||
+ | In the net, face <math>5</math> shares a vertex with the bottom-left corner of face <math>Q</math>, but it doesn't share an edge with face <math>Q</math>, so face <math>5</math> must be the face diagonally across from face <math>Q</math>'s bottom-left corner in the octahedron. Notice that in the octahedron, this face is the only one that doesn't share any vertices with face <math>?</math>. On the net, face <math>5</math> shares at least 1 vertex with all of the other faces except for face <math>1</math>. Thus, the number of face <math>?</math> is \boxed{\text{(A)}~1}. -UnknownMonkey | ||
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==Animated Video Solution== | ==Animated Video Solution== | ||
https://youtu.be/ECqljkDeA5o | https://youtu.be/ECqljkDeA5o |
Revision as of 21:15, 25 January 2023
Contents
Problem
A regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedrons shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of ?
Solution (Intuition)
The answer is Use intuition to bring it down to
guesses
or
and guess from there or you could actually fold the paper.
-apex304
Solution 2
In the net, face shares a vertex with the bottom-left corner of face
, but it doesn't share an edge with face
, so face
must be the face diagonally across from face
's bottom-left corner in the octahedron. Notice that in the octahedron, this face is the only one that doesn't share any vertices with face
. On the net, face
shares at least 1 vertex with all of the other faces except for face
. Thus, the number of face
is \boxed{\text{(A)}~1}. -UnknownMonkey
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using 3D Visualization)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3789
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.